Y=(4n+4)(8n-3)

Solution for Y=(4n+4)(8n-3) equation:

=(4Y+4)(8Y-3)

We move all terms to the left:

-((4Y+4)(8Y-3))=0

We multiply parentheses ..

-((+32Y^2-12Y+32Y-12))=0


We calculate terms in parentheses: -((+32Y^2-12Y+32Y-12)), so:

(+32Y^2-12Y+32Y-12)

We get rid of parentheses

32Y^2-12Y+32Y-12

We add all the numbers together, and all the variables

32Y^2+20Y-12

Back to the equation:

-(32Y^2+20Y-12)


We get rid of parentheses

-32Y^2-20Y+12=0

a = -32; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·(-32)·12
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-44}{2*-32}=\frac{-24}{-64}
=3/8
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+44}{2*-32}=\frac{64}{-64}
=-1
$